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JEE Main 2019 (9 Apr, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2019 (9 Apr, Shift 2)

Q37JEE Main 2019 · 9 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K)

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Answer: (c)

  1. Spring energy =12kx2=12(800)(0.02)2=0.16 J= \tfrac{1}{2}kx^2 = \tfrac{1}{2}(800)(0.02)^2 = 0.16\text{ J}, all turned into heat.
  2. Heat capacity =1×4184+0.5×400=4384 J/K= 1 \times 4184 + 0.5 \times 400 = 4384\text{ J/K}
  3. ΔT=0.164384≈3.6×10−5 K\Delta T = \dfrac{0.16}{4384} \approx 3.6 \times 10^{-5}\text{ K}
  4. So the answer is (c) 10−5 K10^{-5}\text{ K}.
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